A The Electro-cardiology signal (ECG) is a recording of (usually human) heart-ra

Question

A The Electro-cardiology signal (ECG) is a recording of (usually human) heart-rate as a function of time and it is a very popular an valuable diagnostic/health monitoring tool. You wish to design a wearable digital ECG monitor with the following (selected) specifications: Number of ECG channels 2 (so you will record ECG from two locations on the body); Sampling frequency: 200 Hz: Sampling resolution 15 bits (each sample will be represented by 16 digital bits) Assuming that you wish to store up to approximately 3 hours of recording, what minimum amount of memory would you use the monitor? B Assume that the LTI system is described by the following differential equation: 2y(t) - 4 dy(t)/dt + 2 d^3y(t)/dt^3 - dy^4(t)/dt^4 - x(t) + 3 d^2x(t)/dt^2 = 0 where x(t) is the input signal and y(t) is the output signal. The corresponding transfer function is: Consider the discrete-time signal, whose part is illustrated below. What expression(s) would represent this signal?

Explanation / Answer

[A]

Sampling freq.= The number of samples taken per second

So we will have 200 samples per seconds

Sampling Resolution= The number of bits assigned to each sample.

i.e 16 bits

for one sample we require 16 bits;

so for 200 samples we will require 16x200=3200 bits

So we will require 3200 bits to store 1 second of recording.

for 1 second it is 3200 bits

then 60 seconds (1 minute) it will be 3200 x 60 bits

then 60 minutes (1 hr.) it will be 3200 x 60 x 60 bits

then for 3 hr it will be 3200 x 60 x 60 x 3 bits

So we will require 34560000 bits to store 3 hr of recording.

SO FOR TWO CHANNEL IT WIIL BE 34560000 x 2 = 69120000 bits

and we know that

1 BYTE = 8 bits

1 bit = 1/8 BYTE

SO

69120000 bits = 69120000 / 8 BYTES

= 8640000 BYTES

= 8.64 x 106 BYTES

   = 8.64 MegaBytes

So Minimum Amount of Memory required = 8.64 MB

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